## mathcalculus related rate problem: help please. (attached) 8 months ago 8 months ago

1. mathcalculus

2. SithsAndGiggles

Recall the distance formula: $d^2=(x-x_0)^2+(y-y_0)^2$ In this particular problem, you have $$(x_0,y_0)=(0,0)$$ (the origin), since you want to find the distance between any given point on the curve $$(x,y)$$ and the origin. $d^2=x^2+y^2$ Substitute the given equation of the curve: $d^2=x^2+\left(4\sqrt{2x+2}\right)^2\\ d^2=x^2+16(2x+2)\\ d^2=x^2+32x+32$ Now differentiate both sides implicitly with respect to some dummy variable $$t$$: $2d\frac{dd}{dt}=(2x+32)\frac{dx}{dt}$ You're told that at the point $$(1,8)$$, $$x$$ is increasing at a rate of 3 units/s, which translates to $$\dfrac{dx}{dt}=3$$. $2d\frac{dd}{dt}=(2\cdot1+32)(3)\\ 2d\frac{dd}{dt}=102$ You're asked to find $$\dfrac{dd}{dt}$$, so first you need to find $$d$$. To do this, find the distance between $$(1,8)$$ and $$(0,0)$$, then plug it into the above equation. Lastly, solve for $$\dfrac{dd}{dt}$$.

3. mathcalculus

@SithsAndGiggles omg thank you!! :)

4. SithsAndGiggles

Yer welkum

5. mathcalculus

@SithsAndGiggles hey, i found the distance which is 8 right? then i plugged in 8 into the equation and got 51/8... however, something is wrong?

6. satellite73

i get $$\frac{19}{8}$$

7. mathcalculus

oh nvm i got it.

8. mathcalculus

it's 6.32577

9. mathcalculus

to find the distance is this:

10. mathcalculus

i got it... thank you! @satellite73

11. satellite73

$d^2=x^2+32x+32$ $2dd'=2xx'+32x'$ $dd'=xx'+32x'$ $8d'=3+96$

12. satellite73

oh right doh

13. satellite73

distance isn't 8 is it? distance is $$\sqrt{65}$$

14. satellite73

which is pretty close to 8...

15. mathcalculus

yes

16. satellite73

is the final answer $$\frac{99}{\sqrt{65}}$$ ?

17. mathcalculus

close yes... i thought i had to do this: x2-x1/y2-y1

18. mathcalculus

it's 51/sqrt(65)/65

19. satellite73

oh damn i forgot to divide 32 by 2 must be past my bedtime

20. mathcalculus

lol